what a nice functional equation!

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[Calcprof]

This problem is screaming Legendre transform at me -- together with equality conditions for equality in the inequality of a convex function and its convex conjugate (Legendre transform). In fact your proof mirrors the proof that the conjugate of the conjugate is the original function.

[stephenhamer]

Do we need to use calculus to maximize { 2xy - y^2 | y ∊ R }? Surely we can write max { 2xy - y^2 | y ∊ R } = max { x^2 - x^2 + 2xy - y^2 | y ∊ R } = max { x^2 - ( x - y )^2 | y ∊ R } = x^2, which establishes f(x) = x^2 as a solution very directly

[aweebthatlovesmath]

I've seen so many functional equations but this one was really unique!

[Ganerrr]

The first step to these types of problems feel so obvious once you see it but I find really hard to figure out on my own

[Павал-лч]

I like feeding these problems to ChatGPT to make sure I can still have a job

[ВасилийТёркин-кх]

My favourite videos are about functional equations. I always hope that there exists a non trivial solution to the equation in question. It's a bummer when solution doesn't exist or it's f(x) = x. I try to solve equations by myself before watching a video and it was interesting to recall properties of max. I enjoyed this one a lot.

[charleyhoward]

I couldn't have figured this out on my own if my life depended on it

[joelpaddock]

Kind of reminds me of the definition of something called a Legendre transform

[danielleza]

Here's my simple proof:
The beginning is the same as Michael's proof, where we find that f(x) >= x^2 and also that x^2 is a solution.

To prove the uniqueness of this solution, let's assume that another solution exists g(x). We know the inequality g(x) >= x^2, therefore this solution can be written as g(x) = x^2 + h(x) where h(x) is non-negative for all x, and positive for atleast one value of x, call it x=a.

Now we get the following:

x^2 + h(x) = g(x) = max{ 2xy - y^2 - h(y) | y in R} <= max{ 2xy - y^2 | y in R} = x^2

The inequality is due to the fact that h(y) is non-negative, so removing it only increases the maximum.
We got that for all x, x^2 + h(x) <= x^2, but we know that for x=a we have a^2 + h(a) > a^2, contradiction.

[BenfanichAbderrahmane]

Hello Micheal, can you tell me where do you find these type of problems ?

[Alan-zftt]

First of all I'd like to thank Michel and his team wishing them all a wonderful new year 🔔 🔔 🔔 🔔
And then offer thanks for the content of this video.
It was exhilerating

[anshulrajsingh]

Another way to finish:
Suppose that for some y = a the maximum is achieved, then
f(x) = 2xa - f(a) <= 2ax - a² <= x² implying that f(x) = x².

[paulryanlonghas]

In general, f = f^* if and only if f = 1/2|| • ||² where f^* is the Fenchel conjugate of f.

[icewlf]

someone please find an error in my reasoning:

f(0) = max{0-f(y)|y∈ℝ} = -min{f(y)|y∈ℝ} = -m
suppose that for some x, f(x)<0
f(x)<0 ⇔ for every y∈ℝ: 2xy-f(y)<0 ⇔ for every y∈ℝ: 2xy<f(y) label this last statement (§)
(§) implies 0<f(0) implies 0<-m implies m<0
(§) implies both: f(x)<0 and 2x²<f(x), contradiction with f(x)<0, so for all x∈ℝ: f(x)≥0
last two rows also give us a contradiction, because minimum value of f is negative, but it can't be negative as we concluded in the last row

f can't exist

[albertobarreirohermida]

what about f(x)=x^2+c? it achieves the same property if i'm not mistaken

[talastra]

Learning more today, and being confused as usual :)

[ruffifuffler]

2 male worms, minus, a half worm = 3/2 sand worms.

Yes, that's the kind of math suggested by the radical double induction operator acting as an inverter with possible rational confinement, ready to truncate to rational confinement by shedding the entire opposite sex on the division line, at the border between dominance, and subdominance.

Remember, the postman always rings twice?

[stephenl]

Is "super unique" more unique than "unique"?

[kkanden]

not a constant or identity function as the solution! yay!